Propanoic acid is a monoprotic weak acid. This means that at equilibrium, the reaction equation would be:

HA  <--->  H+  +  A-

HA = propanoic acid

A- = conjugate base

H+ = hydronium ions

We can get the value of H+ from the given pH which is 3.25.

pH = -log [H+]

3.25 = -log [H+]

[H+] = 10^(-3.25)

[H+] = 5.6234x10^-4

Next, we draw the ICE table. 

       HA           <--->      H+    +     A-

I       M                           0             0

C     -x                         +x           +x

E   M-x                           x             x

We can see that

x = [H+] = [A-] = 5.6234x10^-4 = 5.62x10^-4

M would be the initial concentration of the acid HA. To solve for that value, we will use the equilibrium expression.

ka = [H+][A-]/ M-X

ka = [x][x]/M-x

ka = 10^(-pka) = [x][x]/M-x

**remember that pka is given in the problem.

10^(-4.874) = (5.6234x10^-4)^2/ M-5.6234x10^-4

by algebra, M = 0.02422 = 0.024

% ionization = X/M *100

% ionization = (5.6234x10^-4/0.02422) x100

% ionization = 2.32%